Auto-Math
Sachant que \(ABCDEF\) est un hexagone régulier inscrit dans un cercle de centre \(O\) , comparez les angles \(\widehat{DBA}\) et \(\widehat{DOA}\).
\( \widehat{DBA}=\widehat{DOA} \)
\( 2\widehat{DBA}=\widehat{DOA} \)
\(\widehat{DBA}=2\widehat{DOA} \)
\( \dfrac{1}{2}\widehat{DBA}=\widehat{DOA} \)
Résolvez l'équation \(\cos(3x+\pi) = \cos x \).
\( S=\left\{-\dfrac{\pi}{2}+k\pi,\, -\dfrac{\pi}{4}+k\dfrac{\pi}{2};\, k\in\mathbb{Z}\right\} \)
\( S=\left\{-\dfrac{\pi}{2},\, -\dfrac{\pi}{4};\, k\in\mathbb{Z}\right\} \)
\( S=\left\{-\dfrac{\pi}{2}+k\pi;\, k\in\mathbb{Z}\right\} \)
\( S=\left\{-\dfrac{\pi}{2}+2k\pi,\, -\dfrac{\pi}{4}+2k\dfrac{\pi}{2};\, k\in\mathbb{Z}\right\} \)
Déterminez à l'aide du cercle trigonométrique la valeur de \(\cos\dfrac{11\pi}{6} \).
\( \dfrac{1}{2} \)
\( -\dfrac{1}{2} \)
\( \dfrac{\sqrt{3}}{2} \)
\( -\dfrac{\sqrt{3}}{2} \)
Résolvez l'équation \(\sin 2x = \sin \dfrac{\pi}{4}\) .
\( S=\left\{\dfrac{\pi}{8}\right\} \)
\( S=\left\{\dfrac{\pi}{8},\, \dfrac{3\pi}{8}\right\} \)
\(S=\left\{\dfrac{\pi}{8}+2k\pi,\, \dfrac{3\pi}{8}+2k\pi;\, k\in\mathbb{Z}\right\} \)
\( S=\left\{\dfrac{\pi}{8}+k\pi,\, \dfrac{3\pi}{8}+k\pi;\, k\in\mathbb{Z}\right\} \)
Donnez la valeur de \( \sin {\pi \over 2}\) .
1
-1
0
90
Résolvez l'équation \(\sin x = \sin 2x \).
\( S=\left\{k\pi,\, \dfrac{\pi}{3}+2k\pi,\, \dfrac{5\pi}{3}+2k\pi;\, k\in\mathbb{Z}\right\} \)
\( S=\left\{0,\, \pi,\, \dfrac{\pi}{3},\, \dfrac{5\pi}{3}\right\} \)
\( S=\left\{k\pi,\, \dfrac{\pi}{3}+2k\pi;\, k\in\mathbb{Z}\right\} \)
\( S=\left\{k\pi,\, \dfrac{\pi}{6}+2k\pi,\, \dfrac{11\pi}{6}+2k\pi;\, k\in\mathbb{Z}\right\} \)
Résolvez l'équation \(2\sin{3x}+\sqrt{2}=0\) .
\( S=\left\{\dfrac{5\pi}{12},\, \dfrac{7\pi}{12}\right\} \)
\( S=\left\{\dfrac{5\pi}{12}+2k\dfrac{\pi}{3},\, \dfrac{7\pi}{12}+2k\dfrac{\pi}{3};\, k\in\mathbb{Z}\right\} \)
\(S=\left\{\dfrac{5\pi}{12}+2k\pi,\, \dfrac{7\pi}{12}+2k\pi;\, k\in\mathbb{Z}\right\} \)
\( S=\left\{\dfrac{7\pi}{12}+2k\dfrac{\pi}{3};\, k\in\mathbb{Z}\right\} \)
Déterminez à l'aide du cercle trigonométrique la valeur de \(\cos\dfrac{4\pi}{3}\) .
Convertissez en radians l'angle \(390^\circ \).
\(30\mbox{ radians}\)
\(\dfrac{\pi}{3} \mbox{ radians}\)
\( \dfrac{\pi}{6}\mbox{ radians}\)
\( 2\pi \mbox{ radians}\)
Si \(\alpha=53^{\circ}\) , alors le supplémentaire de \(\alpha\) vaut
\(37^{\circ} \)
\( 127^{\circ} \)
\( 233^{\circ} \)
\( 413^{\circ} \)
